Answer:
0.480 grams
Explanation:
Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)
1 : 3 : 1 : 3
Number of moles (n) = Mass in gram/ Molar Mass
Mass of ND₃ = 160 mg
= 0.16 g
Molar mass of ND₃= [14 + (3 x 2.014 )]
= 14 + 6.042
= 20.042 g/mol
Number of moles of ND₃ = 0.16/20.042
= 0.007983 moles
From the reaction equation, the mole ratio between Heavy water (D₂O ) and ND₃ is 3: 1.
This implies that the number of moles of Heavy water (D₂O ) required
= 3 x 0.007983 moles
= 0.023949 moles
Molar mass of Heavy water (D₂O )= [(2.014 x 2) + 16]
= 20.028 g/mol
Mass in grams of Heavy water (D₂O )= Number of moles x Molar mass
= 0.023949 x 20.028
= 0.4797 grams
≈ 0.480 grams
Answer:
Each orbit has a specific energy level.
Would you mind marking it the brainliest:).
Answer:
endoplasmic reticulum (ER)
Answer:
<h2>1.54 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>1.54 mL</h3>
Hope this helps you
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
